This used to be about my music Now it is just overflow from other places.

Friday, May 9, 2014

Deriving the additive properties of the integers from their multiplicative properties

Usually formal arithmetic begins with
additive properties, such as 0 and the +1 operator. Along the way,
multiplication n*m gets defined as the successive addition of n, m
times. This leads to the notion of primes and the topic of the
distribution of primes. Instead
I want to look at the integers in terms of their multiplicative
properties and see if additive properties (in particular the standard
ordering of the positive integers) can be derived. So begin with a set
of prime numbers, given in order 1,2,3,...... with 1 as multiplicative
identity and other numbers formed by stringing together primes with
multiplications. The question is:how might these these composites be ordered? So here is a very incomplete thought.

We will try to derive an ordering using the symbol “n” to
mean immediate neighbor, so that if B is the immediate neighbor to the right of
A we can write:

A n B

If there is a sequence of zero of more numbers C1,C2,C3, …
such that AnC1nC2nC3……..nB then we write:

The relation ‘n’ and the
relation‘nn’ are no not
reflexive (A n A is false, A nn A is false)

The relation ‘n’ is not
transitive (but ‘nn’ is transitive, as noted above)

If A n B then C*A nn C*B (and not C*A n C*B unless C==1)

[added] Every A has a B such that A n B

Key Axioms

(ISOPERIMETRY) If A n B
and C n D where A nn C we must have A*D nn B*C

(PARSIMONY) Composites
occur as early in the sequence as possible without violating (4)

Assume

1 nn 2 nn 3 nn 5 etc.

Axiom (4) is like the isoperimetric concept: of all
rectangles with the same perimeter, the square is the one with the largest
area. Axiom (5) says that composites are packed as closely together as possible;
alternatively, that primes are introduced as infrequently as possible.

The conventional additive order of the (composite) integers
can be derived from these axioms and the order of the primes.

Proof of the partial theorem is something like this: 1 has a
neighbor but it cannot be a composite using just 1, so it must be 2, hence 1 n 2.
Now axiom (5) says we should try to put
2*2 as soon as possible after 2 but that would give us1 n 2 n 2*2.

If we apply axiom (4) this says that 1*(2*2) nn 2*2 which is false by
axiom (1). Hence we need another prime next to 2, call it “3” so we have
1 n 2 n 3. Now we must have 3 n 2*2 because of axiom (5), so we have 1 n
2 n 3 n 2*2. We cannot have 2*2 n 2*3 without violating (4) and all
other composites are even larger than 2*3, so we must have another prime
“5” after 2*2. Now we have 1 n 2 n 3 n 2*2 n5. (The argument continues???)

Not only that, multiplication assumes we can count acts of counting. So counting 1, 2, ..., n as many as m times, give n*m.If you felt like disagreeing, you could ask: how do we guarantee, while counting 'acts of counting' that each one was done in the same way?

Also mention this: if for example the additive order of the composites cannot be derived from the primes, it leads to the question of whether there could be more than one ordering of the composites, using the same primes.

I can now get up to 13 with order guaranteed by the isoperimetry requirement. But there is no good reason I can see for 14<15 unless you start analyzing the space between 5 and 7, where 6 screws you up and adds to complexity in a way that might still come under isoperimetry but is beyond ME.

I was playing with powers of 2 and 3, wondering if there was any clear way to order them using isoperimetry. But there may be no way to order them in any case. Speculate that the only set of primes that lead to order-able composites is the old familiar infinite set of all primes.

But how cool would it be for there to exist a set of -say- 63 primes whose composites could be ordered and form an abelian group under the "next" operation? It would be fun - who doesn't like a new number system?